Two hockey players of opposite teams, while trying to hit a hockey ball on the ground collide and immediately become entangled. One has a mass of 60 kg and was moving with a velocity 5.0 m s–1 while the other has a mass of 55 kg and was moving faster with a velocity 6.0 m s–1 towards the first player. In which direction and with what velocity will they move after they become entangled? Assume that the frictional force acting between the feet of the two players and ground is negligible.

Solution:

Let the first player be moving from left to right. By convention left to right is taken as the positive direction and thus
right to left is the negative direction . If symbols m and u represent the mass and initial velocity of the two
players, respectively. Subscripts 1 and 2 in these physical quantities refer to the two hockey players. Thus,
m1 = 60 kg; u1 = + 5 m s-1; and m2 = 55 kg; u2 = – 6 m s-1
.
The total momentum of the two players before the collision
= 60 kg × (+ 5 m s-1) + 55 kg × (– 6 m s-1)
= – 30 kg m s-1 If v is the velocity of the two entangled players after the collision, the total momentum then
= (m1 + m2 ) × v = (60 + 55) kg × v m s–1
= 115 × v kg m s–1
.
Equating the momenta of the system before and after collision, in accordance with the law of conservation of
momentum, we get
v = – 30/115 = – 0.26 m s–1
.
Thus, the two entangled players would move with velocity 0.26 m s–1 from right to left, that is, in the direction the
second player was moving before the collision.

A collision of two hockey players: (a) before collision and (b) after collision

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